Answer to Question #194255 in Calculus for Maureen

Question

Find the slope of the tangent line at the incident point

3e to the power x = xy+xsquared+ysquared at (0,1)

Write an equation of the tangent line

Answer 1

$$(3e)^x=xy+x^2+y^2$$

Differentiate both parts with respect to $x$ . Use the Chain Rule

$$\dfrac{d}{dx}((3e)^x )=\dfrac{d}{dx}(xy+x^2 +y^2)$$

$$(\ln3+1)(3e)^x=y+x\dfrac{dy}{dx}+2x+2y\dfrac{dy}{dx}$$

Solve for $\dfrac{dy}{dx}$

$$\dfrac{dy}{dx}=\dfrac{(\ln3+1)(3e)^x-2x-y}{x+2y}$$

Find the slope of the tangent line at point $(0, 1)$

$$slope=m=\dfrac{dy}{dx}|_{(0,1)}$$

$$=\dfrac{(\ln3+1)(3e)^0-2(0)-1}{0+2(1)}=\dfrac{\ln3}{2}$$

The equation of the tangent line in point-slope form

$$y-1=\dfrac{\ln3}{2}(x-0)$$

The equation of the tangent line in slope-intercept form

$$y=\dfrac{\ln3}{2}\cdot x+1$$